HELP I WILL GIVE BRAINLIEST!

an engine has been design to work between source and the sink at temperature 177 degree Celsius and 27 degree Celsius respective

irina [24]

Given data

Source temperature (T₁) = 177°C = 177+273 = 450 K

Sink temperature (T₂) = 27°C = 27+273 = 300 K

Energy input (Q₁) = 3600 J ,

Work done = ?

We know that, efficiency (η) = Net work done ÷ Heat supplied

η = W ÷ Q₁

W = η × Q₁

First determine the efficiency ( η ) = ?

Also, we know that ( η ) = (T₁ - T₂) ÷ (T₁)

= 33.3% = 0.333

Now, Work done is W = η × Q₁

= 0.33 × 3600

W = 1188 J

Work done by the engine is 1188 J

4 0

3 years ago

De donde eres responde y te doy corona

irakobra [83]

soy de texas, united states

8 0

2 years ago

A refrigeration cycle has Qout = 1000 Btu and Wcycle = 300 Btu. Determine the coefficient of performance for the cycle.

DENIUS [597]

Answer:

The coefficient of performance for the cycle is 2.33.

Explanation:

Given that,

Output energy $Q_{out}=1000\ Btu$

Work done $W_{cycle}=300\ Btu$

We need to calculate the coefficient of performance

Using formula of the coefficient of performance

$COP=\dftrac{Q_{in}}{W_{cycle}}$

We need to calculate the $Q_{in}$

$W_{cycle}=Q_{out}-Q_{in}$

Put the value into the formula

$300=1000-Q_{in}$

$Q_{in}=300-1000$

$Q_{in}=700\ Btu$

Now put the value of $Q_{in}$ into the formula of COP

$COP=\dfrac{700}{300}$

$COP=\dfrac{7}{3}=2.33$

Hence, The coefficient of performance for the cycle is 2.33.

5 0

3 years ago

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

or

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

or

$cos X=0.8951$

or

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

or

$cos Y=0.649$

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

3 years ago

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago